Question

我正在显示一个页面，其中包含productname，productdetails，product image等所有产品详细信息。单个产品有多种类型。首先，只应显示一个产品。根据类型动态创建单选按钮。单击单选按钮并选择其他类型的图像时，必须更改productcolor。我已经传递了值$ color和productcolor相应的更改。 1.我们如何改变图像？我们如何使用img_id检索img？ 2.选择数量后，当我发布cart.php时，数量值在cart.php中返回为0.为什么会发生？

        <table width="1000" border="0" cellspacing="0" cellpadding="0">
        <tr>
            <td width="300" height="600" align ="center" ><img id ="img" src="inventory_images/<?php echo $img_id; ?>.jpg" width="300" height="400" alt=""/></td>
            <br>

        <p id = "test"><?php echo $img_id; ?></p>

            <td width="150" height = "400" align="center"  >

             <p align="left"><?php echo $name; ?></p>
             <p align ="left" id = "demo"><?php echo $color; ?></p>
             <p align = "left"><?php echo $product_detail; ?></p>

             <p align = "left">Price : $<?php echo $product_price; ?></p>


              <?php while($rows = mysqli_fetch_array($result))
    {
        $color = $rows["clr"];
        $rimg_id = $rows["img_id"];



          ?>      



<ul align="left" id="menu">
<li><p onclick="myFunction('<?php echo $color; ?>' ,'<?php echo $rimg_id; ?>');"><input   type ="radio" name="radio" id ="php" value="<?php echo $color; ?>" >
<?php echo $color; ?></p>
</li>
</ul>


 <?php } ?>

<script type="text/javascript" >
function myFunction(a,b) {


       document.getElementById("demo").innerHTML = a;
    document.getElementById("test").innerHTML = b;
    document.getElementById("imgid").value = b;
    //document.getElementById("img").src= "inventory_images/.jpg";
       //how to get my images stored in my folder??



}
</script>

 <p align = "left" style="color:#33F">Quantity</p> <input type='button' name='add' onclick='javascript: document.getElementById("qty").value++;' value='+'/>
<input type='text' name='qty' id='qty' value = '1' size = '1' />
</div><input type='button' name='subtract' onclick='javascript: subtractQty();' value='-'/>
<script type="text/javascript">
        function subtractQty(){
            if(document.getElementById("qty").value - 1 < 1)
                return;
            else
                 document.getElementById("qty").value--;
        }
        </script>




              <form id = "add" name = "Add to cart" method = "POST" action = "cart.php" >
             <p><input type = "hidden" id="imgid" name = "imgid" value =""><p>  
              <p align="left"><input type = "submit" name = "submit" id = "submit"  value = "Add to Cart" ><p>
        </form>

Answer 1

您必须为每种类型创建一个p元素。坚持到底。

<div ng-app='app'>
    <div ng-controller='MainController'>
        <div ng-bind='getName()'></div>

        <div ng-controller='Child1'>
            <div ng-bind='getName()'></div>

            <div ng-controller='Child2'>
                <div ng-bind='getName()'></div>
            </div>
        </div>
    </div>
</div>

您必须在图像src加载器中连接函数参数b：

<ul align="left" id="menu">
<?php while($rows = mysqli_fetch_array($result))
{
    $color = $rows["clr"];
    $rimg_id = $rows["img_id"];
?>     
<li>
<p onclick="myFunction('<?php echo $color; ?>' ,'<?php echo $rimg_id; ?>');">
<input type ="radio" name="radio" value="<?php echo $color; ?>">
<?php echo $color; ?>
</p>
</li>
<?php } ?>
</ul>

从动态javascript onclick函数获取图像路径

1 个答案: